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Semenov [28]
2 years ago
11

3. How does a carbon-14 atom change back into a nitrogen atom?

Chemistry
2 answers:
julsineya [31]2 years ago
5 0

<em>Carbon 14 change back into the stable nuclei Nitrogen by </em><em>emitting electron and electron antineutron.</em>

<u>Explanation:-</u>

Carbon 14 is one of the radioactive isotope elements that contains 6 protons and 8 neutron elements.

A general chemical reaction of Carbon 14 is given below.

14_{6} C \rightarrow 14_{7} N+B e t a^{-1}

This above reaction can be categorized under Nuclear Reaction. When atomic number 6 (Carbon) was irradiated, Beta ray emission happen and new stable nuclei with atomic number 7 (Nitrogen) formed without change in atomic mass of any nuclei. As well as some amount of energy also emitted through this reaction.

madreJ [45]2 years ago
4 0

Answer:  an electron and an electron antineutrino, one of the neutrons in the carbon-14 atom decays to a proton and the carbon-14 (half-life of 5,700 ± 40 years) decays into the stable (non-radioactive) isotope nitrogen-14.

Neutrons: 8

Isotope mass: 14.003241 u

Symbol names: 14C; carbon-14, C-14, radioca...

Protons: 6

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Which of the following has the same empirical and molecular formula?
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NaCl is ___ formula of common salt
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4 0
3 years ago
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A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?
Viktor [21]

The answer for the following problem is mentioned below.

  • <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules.</em></u>

Explanation:

Given:

mass of calcium phosphate (Ca_{3}(PO_{4} )_{2} ) = 125.3 grams

We know;

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = 120 + 3(95)

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} )  = 120 +285 = 405 grams

<em>We also know;</em>

No of molecules at STP conditions(N_{A}) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

\frac{m}{M} =\frac{N} }{}N÷N_{A}

\frac{405}{125.3} =\frac{N}{6.023*10^23}

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules</em></u>

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3 years ago
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